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malfutka [58]
3 years ago
13

What type of weather would be in the middle of the picture, where the barometric pressure is 1000 millibars?

Physics
1 answer:
UkoKoshka [18]3 years ago
4 0
It would be D. Hope that helped :)
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a deer with a mass of 176 kg is running head-on towards you with a velocity of 19 m/s. you are going north. find the magnitude a
Igoryamba

Momentum = Mass × Velocity

According to this formula,

Momentum of deer = 176 × 19 = 3344 kg•m/s.

Since you are heading north and the deer is running towards you, the direction of the deer' s momentum is north as well.

6 0
3 years ago
Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kΩ, and
SOVA2 [1]

Answer:

10042.6 ohm

Explanation:

f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

C = 5 nF = 5 x 10^-9 F

XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2  + (XL - Xc)^2

Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2

Z = 10042.6 ohm

4 0
3 years ago
How much heat energy in megajoules is needed to convert 7 kilograms of ice at -9c to water at 0c
Musya8 [376]
The latent heat of fusion of water is 344 j/p.This will take 344x300 or 100200 J to melt the ice. once it melted its at oc
6 0
3 years ago
Read 2 more answers
Ben Rushin is waiting at a stop light. Turns green, ben accelerated from rest at a rate of 6.00 m/s squared for a time of 4.10 s
Lera25 [3.4K]
D= vt +.5at^2
since he started at rest, v (initial velocity) is 0
so d=.5at^2
d = .5 (6m/s^2) (4.1s)^2 
then put that into a calculator.

4 0
3 years ago
A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi
Elis [28]

Answer: 115.52\ N

Explanation:

Given

Length of plank is 1.6 m

Force F_1 is applied on the left side of plank

Force F_2 is applied 43 cm from the left end O.

Mass of the plank is m=13.7\ kg

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N

Net vertical force must be zero on the plank

\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N

8 0
2 years ago
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